3.2 Quadratic Simultaneous Equations
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Learning Objectives
- Understand what quadratic simultaneous equations are
- Solve a system with one linear equation and one quadratic equation
- Solve a system with two quadratic equations
- Interpret the geometric meaning of solutions
- Apply quadratic simultaneous equations to solve real-world problems
Key Concepts
Introduction to Quadratic Simultaneous Equations
Quadratic simultaneous equations involve systems where at least one of the equations is quadratic (contains terms with variables raised to the power of 2).
The most common types are systems with one linear and one quadratic equation, or systems with two quadratic equations.
Solutions to these systems can be found algebraically or interpreted geometrically as the points of intersection between curves.
Solving a System with One Linear and One Quadratic Equation
The most straightforward approach is to use the substitution method.
First, rearrange the linear equation to express one variable in terms of the other.
Then substitute this expression into the quadratic equation to obtain a quadratic equation in one variable.
Solve this quadratic equation and then substitute the solutions back to find the corresponding values of the other variable.
Example 1
Solve the system of equations: y = x + 1 x² + y² = 5
Solution:
Step 1: From the first equation, we have y = x + 1. Step 2: Substitute this into the second equation. x² + (x + 1)² = 5 x² + x² + 2x + 1 = 5 2x² + 2x + 1 = 5 2x² + 2x - 4 = 0 x² + x - 2 = 0 Step 3: Solve this quadratic equation using the quadratic formula or factorization. x² + x - 2 = 0 (x + 2)(x - 1) = 0 x = -2 or x = 1 Step 4: Substitute these values back to find the corresponding y-values. For x = -2: y = -2 + 1 = -1 For x = 1: y = 1 + 1 = 2 So the solutions are (-2, -1) and (1, 2).
Example 2
Solve the system of equations: 2x + y = 10 x² + y² = 34
Solution:
Step 1: From the first equation, express y in terms of x. 2x + y = 10 y = 10 - 2x Step 2: Substitute this into the second equation. x² + (10 - 2x)² = 34 x² + 100 - 40x + 4x² = 34 5x² - 40x + 100 = 34 5x² - 40x + 66 = 0 x² - 8x + 13.2 = 0 Step 3: Solve this quadratic equation using the quadratic formula. x = (8 ± √(64 - 4 × 1 × 13.2)) / 2 x = (8 ± √(64 - 52.8)) / 2 x = (8 ± √11.2) / 2 x ≈ (8 ± 3.35) / 2 x ≈ 5.67 or x ≈ 2.33 Step 4: Substitute these values back to find the corresponding y-values. For x ≈ 5.67: y ≈ 10 - 2(5.67) ≈ 10 - 11.34 ≈ -1.34 For x ≈ 2.33: y ≈ 10 - 2(2.33) ≈ 10 - 4.66 ≈ 5.34 So the solutions are approximately (5.67, -1.34) and (2.33, 5.34).
Solving a System with Two Quadratic Equations
Systems with two quadratic equations are generally more complex to solve.
One approach is to try to eliminate one of the squared terms by subtracting one equation from the other, if possible.
Another approach is to solve one equation for one variable in terms of the other, and then substitute into the second equation.
These systems can have up to four solutions, corresponding to the possible intersections of two conic sections.
Example 1
Solve the system of equations: x² + y² = 25 x² - y² = 7
Solution:
Step 1: Add the two equations to eliminate y². x² + y² = 25 x² - y² = 7 -------------- 2x² = 32 x² = 16 x = ±4 Step 2: Substitute these values back into the first equation to find y. x² + y² = 25 16 + y² = 25 y² = 9 y = ±3 Step 3: Form the solution pairs. Since both x and y can be positive or negative, we have four possible solutions. (4, 3), (4, -3), (-4, 3), and (-4, -3) Step 4: Verify these solutions by checking that they satisfy both original equations. For (4, 3): 4² + 3² = 16 + 9 = 25 ✓ and 4² - 3² = 16 - 9 = 7 ✓ For (4, -3): 4² + (-3)² = 16 + 9 = 25 ✓ and 4² - (-3)² = 16 - 9 = 7 ✓ For (-4, 3): (-4)² + 3² = 16 + 9 = 25 ✓ and (-4)² - 3² = 16 - 9 = 7 ✓ For (-4, -3): (-4)² + (-3)² = 16 + 9 = 25 ✓ and (-4)² - (-3)² = 16 - 9 = 7 ✓ All four solutions satisfy both equations, so the solutions are (4, 3), (4, -3), (-4, 3), and (-4, -3).
Example 2
Solve the system of equations: x² + y² = 10 x² + y = 6
Solution:
Step 1: From the second equation, express x² in terms of y. x² = 6 - y Step 2: Substitute this into the first equation. (6 - y) + y² = 10 6 - y + y² = 10 y² - y - 4 = 0 Step 3: Solve this quadratic equation using the quadratic formula or factorization. y² - y - 4 = 0 (y - 2)(y + 2) = 0 y = 2 or y = -2 Step 4: Substitute these values back to find the corresponding x-values. For y = 2: x² = 6 - 2 = 4, so x = ±2 For y = -2: x² = 6 - (-2) = 8, so x = ±2√2 So the solutions are (2, 2), (-2, 2), (2√2, -2), and (-2√2, -2).
Geometric Interpretation of Solutions
Geometrically, the solutions to a system of simultaneous equations represent the points of intersection of the curves defined by the equations.
A linear equation represents a straight line, while a quadratic equation typically represents a conic section such as a circle, ellipse, parabola, or hyperbola.
The number of solutions corresponds to the number of intersection points between the curves.
Example 1
Interpret geometrically the solutions to the system: y = x² y = 2x
Solution:
Step 1: Identify the curves represented by each equation. y = x² represents a parabola with vertex at the origin, opening upward. y = 2x represents a straight line passing through the origin with slope 2. Step 2: Find the points of intersection algebraically. Substituting the second equation into the first: 2x = x² x² - 2x = 0 x(x - 2) = 0 x = 0 or x = 2 For x = 0: y = 2(0) = 0 For x = 2: y = 2(2) = 4 So the points of intersection are (0, 0) and (2, 4). Step 3: Interpret geometrically. The parabola and the line intersect at two points: the origin (0, 0) and the point (2, 4). At the origin, the line is tangent to the parabola (both curves have the same slope at this point). At the point (2, 4), the line crosses through the parabola.
Example 2
Interpret geometrically the solutions to the system: x² + y² = 25 x - y = 1
Solution:
Step 1: Identify the curves represented by each equation. x² + y² = 25 represents a circle centered at the origin with radius 5. x - y = 1 represents a straight line with slope 1 and y-intercept -1. Step 2: Find the points of intersection algebraically. From the second equation: y = x - 1 Substituting into the first equation: x² + (x - 1)² = 25 x² + x² - 2x + 1 = 25 2x² - 2x - 24 = 0 x² - x - 12 = 0 (x - 4)(x + 3) = 0 x = 4 or x = -3 For x = 4: y = 4 - 1 = 3 For x = -3: y = -3 - 1 = -4 So the points of intersection are (4, 3) and (-3, -4). Step 3: Interpret geometrically. The circle and the line intersect at two points: (4, 3) and (-3, -4). These are the only two points that lie on both the circle and the line simultaneously.
Applications of Quadratic Simultaneous Equations
Quadratic simultaneous equations arise in many real-world problems, particularly those involving areas, distances, or optimization.
Examples include problems about the dimensions of shapes with known area and perimeter, projectile motion, and economic models.
The key to solving these problems is to identify the variables and set up the appropriate equations based on the given information.
Example 1
A rectangular garden has a perimeter of 26 meters and an area of 40 square meters. Find the dimensions of the garden.
Solution:
Step 1: Define the variables. Let x be the length and y be the width of the garden, both in meters. Step 2: Set up the equations based on the given information. Perimeter: 2x + 2y = 26 Area: xy = 40 Step 3: Simplify the first equation. 2x + 2y = 26 x + y = 13 Step 4: Express y in terms of x. y = 13 - x Step 5: Substitute this into the area equation. xy = 40 x(13 - x) = 40 13x - x² = 40 x² - 13x + 40 = 0 Step 6: Solve this quadratic equation using the quadratic formula or factorization. x² - 13x + 40 = 0 (x - 8)(x - 5) = 0 x = 8 or x = 5 Step 7: Find the corresponding values of y. For x = 8: y = 13 - 8 = 5 For x = 5: y = 13 - 5 = 8 So the dimensions of the garden are either 8 meters by 5 meters or 5 meters by 8 meters, which are the same rectangle.
Example 2
Two positive numbers have a sum of 15 and a product of 56. Find these numbers.
Solution:
Step 1: Define the variables. Let x and y be the two positive numbers. Step 2: Set up the equations based on the given information. Sum: x + y = 15 Product: xy = 56 Step 3: Express y in terms of x. y = 15 - x Step 4: Substitute this into the product equation. xy = 56 x(15 - x) = 56 15x - x² = 56 x² - 15x + 56 = 0 Step 5: Solve this quadratic equation using the quadratic formula or factorization. x² - 15x + 56 = 0 (x - 8)(x - 7) = 0 x = 8 or x = 7 Step 6: Find the corresponding values of y. For x = 8: y = 15 - 8 = 7 For x = 7: y = 15 - 7 = 8 So the two numbers are 7 and 8.
Special Cases and Considerations
Some systems of quadratic equations may have no real solutions, which geometrically means the corresponding curves do not intersect in the real plane.
Other systems may have infinitely many solutions, which occurs when the curves coincide (e.g., two identical circles).
When solving applied problems, it's important to check that the solutions make sense in the context of the problem (e.g., negative lengths are usually not meaningful for physical dimensions).
Example 1
Determine the number of solutions to the system: x² + y² = 1 x² + y² = 4
Solution:
Step 1: Analyze the equations. The first equation, x² + y² = 1, represents a circle centered at the origin with radius 1. The second equation, x² + y² = 4, represents a circle centered at the origin with radius 2. Step 2: Determine if the system is consistent. Subtracting the first equation from the second: x² + y² - (x² + y²) = 4 - 1 0 = 3 This is a contradiction, which means the original system of equations is inconsistent. Wait, that's not right. Let's try again. The two equations represent two different circles with the same center. Since they have different radii (1 and 2), they don't intersect and don't share any points. So the system has no real solutions.
Example 2
Determine the number of solutions to the system: x² + y² = 4 x² + y² = 4
Solution:
Step 1: Analyze the equations. Both equations represent the same circle: a circle centered at the origin with radius 2. Step 2: Determine the number of solutions. Since both equations represent exactly the same set of points, any point that satisfies one equation also satisfies the other. So the system has infinitely many solutions, which can be expressed as the set of all points (x, y) such that x² + y² = 4.
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